Question: Complete the square to solve for $x$. $x^{2}+12x+32 = 0$
Answer: Begin by moving the constant term to the right side of the equation. $x^2 + 12x = -32$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $12$ , half of it would be $6$ , and squaring it gives us ${36}$ $x^2 + 12x { + 36} = -32 { + 36}$ We can now rewrite the left side of the equation as a squared term. $( x + 6 )^2 = 4$ Take the square root of both sides. $x + 6 = \pm2$ Isolate $x$ to find the solution(s). $x = -6\pm2$ So the solutions are: $x = -4 \text{ or } x = -8$ We already found the completed square: $( x + 6 )^2 = 4$